# $L^p$ estimates for $v\in W_0^{2,p}(\Omega)$ (Elliptic equations)

I am reading Gilbarg/Trudinger Chapter 9. Now on Theorem 9.11 page 235 reprint of the 1998 edition. And as usual I am stuck in on of the assertions of the proof.

It says that given $\Omega\subset \mathbb{R}^n$ bounded domain, and for every $v\in W_0^{2,p}(\Omega)$, $1<p<\infty$, on can obtain the estimate \begin{equation*} ||D^2v||_{p;\Omega}\leq \frac{C}{\lambda}||L v||_{p,\Omega} \quad \text{with } C=C(n,p) \end{equation*}

Where $Lv := a^{ij}D_{ij}v$ (Einstein summation convention) is a uniformly elliptic differential operator with constant coefficients and $\lambda$ is the constant such that $a^{ij}\xi_i\xi_j\geq\lambda|\xi|^2$.

In order to prove this the book reccomends to use two previous results,

1) Under the same hypothesis on $v$ be have the estimate (Corollary 9.10)
\begin{equation*} ||D^2v||_{p;\Omega}\leq C||\Delta v||_{p,\Omega} \end{equation*}

2) One can use a linear transformation $Q$ (This idea comes from Lemma 6.1 in the book) to define a new function $\tilde{v}$ such that \begin{equation*} a^{ij}D^{ij}v(x) = \tilde{a}^{ij}\tilde{v}(y) \end{equation*} where $y=xQ$.

I tried to use this two results by taking $\tilde{v}$ such that $\tilde{a}^{ij}\tilde{v}(y)=\Delta \tilde{v}$ and then seeing how are the relations between the norms of $v$ and $\tilde{v}$ involved in the estimate but could not get to any successful calculation. I'll add those calculations if necessary. But if you have a hint on a different, successful, approach it is welcomed.

• This is the Calderon zygmund inequality my friend it is not a easy joke:) – Guy Fsone Nov 10 '17 at 19:22
• What is $L_0$? specify that – Guy Fsone Nov 10 '17 at 19:25
• @GuyFsone I changed the notation to simply $L$ because the subindex was unnecesary. It is a constant coefficients operator. It is used to prove the general non coefficient estimate. The part I don't get is this one exactly, that apparently is not so difficult since is just the constant coefficients case. It is assumed that we have the $L^p$ bounds for the Newtonian potential (That is the same bound that we want but in the special case in which our $L$ is just the Laplacian). This assumed result is what I believe you called the Calderon-Zygmund inequality and Corollary 9.10 in the book. – I.C. Nov 11 '17 at 6:22
• I edited the question so the previous results are easily noted. They can be assumed and used. I think I do understand the demonstrations of those results. – I.C. Nov 11 '17 at 6:24

Lemma 6.1 describes in detail how to find a linear change of coordinates $y=xQ$ so that $\bar v(y) = v(x(y))$ satisfies $$(\Delta \bar v)(y) = (Lv)(x(y)),$$ so that the estimate for the Laplacian gives $$\|D^2 \bar v \|_p \le C(n,p) \| \Delta \bar v \|_p.\tag{1}$$
We can now rewrite both sides in terms of $v$ to obtain the desired estimate. The RHS is easy: taking the $p^\text{th}$ power for convenience and applying the change of variables formula for integration we find
$$\| \Delta \bar v \|_p^p = \int |\Delta \bar v(y)|^p dy=\int |Lv(x)|^p\left| \det \frac{dy}{dx}\right|dx = \left|\det Q\right| \|L v \|^p_p.\tag{2}$$
The LHS is a little trickier, since in addition to the volume transformation we need to turn the $y$ derivatives into $x$ derivatives. From $y = xQ$ we find $D_{x_i} = Q_{ij}D_{y_j},$ so $|D_x^2 v| \le C(n)|Q|^2|D_y^2 \bar v|.$ Recalling (from 6.1) that $|Q| \le C(n) \lambda^{-1/2}$ and doing the change of variables we find $$\| D^2 v \|_p^p = \int |D_x^2 v|^p dx \le \left|\det Q\right|^{-1}\lambda^{-p}\|D^2 \bar v\|_p^p;$$ so combining with $(1)$ and $(2)$ we find $$\|D^2 v\|^p_p \le \lambda^{-p}C(n,p)\|Lv\|_p^p$$ as desired. (I've been vague about which matrix norm I'm using here, but it doesn't matter - they're all equivalent with constant depending only on $n$.)